The plane ax + by + cz = 1 meets the coordina | vedclass

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(D) The equation of the plane is $ax + by + cz = 1$.To find the intersection with the $x$-axis,set $y = 0$ and $z = 0$,which gives $ax = 1$,so $x = \f

Vedclass · Accepted Answer

$(\frac{1}{3a}, \frac{1}{3b}, \frac{1}{3c})$

Vedclass · Answer

(D) The equation of the plane is $ax + by + cz = 1$.To find the intersection with the $x$-axis,set $y = 0$ and $z = 0$,which gives $ax = 1$,so $x = \frac{1}{a}$. Thus,point $A = (\frac{1}{a}, 0, 0)$.To find the intersection with the $y$-axis,set $x = 0$ and $z = 0$,which gives $by = 1$,so $y = \frac{1}{b}$. Thus,point $B = (0, \frac{1}{b}, 0)$.To find the intersection with the $z$-axis,set $x = 0$ and $y = 0$,which gives $cz = 1$,so $z = \frac{1}{c}$. Thus,point $C = (0, 0, \frac{1}{c})$.The centroid of a triangle with vertices $(x_1, y_1, z_1)$,$(x_2, y_2, z_2)$,and $(x_3, y_3, z_3)$ is given by $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3})$.Substituting the coordinates of $A, B$,and $C$:Centroid $= (\frac{\frac{1}{a} + 0 + 0}{3}, \frac{0 + \frac{1}{b} + 0}{3}, \frac{0 + 0 + \frac{1}{c}}{3}) = (\frac{1}{3a}, \frac{1}{3b}, \frac{1}{3c})$.Therefore,the correct option is $D$.

The plane $ax + by + cz = 1$ meets the coordinate axes at points $A, B$,and $C$. Find the centroid of the triangle $ABC$.

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